Leetcode 1026: Maximum Difference Between Node and Ancestor
You are given the root of a binary tree. Your task is to find the maximum absolute difference between the values of two nodes, where one node is an ancestor of the other. Specifically, you need to find the largest value of |a.val - b.val|, where node a is an ancestor of node b.
Problem
Approach
Steps
Complexity
Input: The input consists of the root of a binary tree, where each node contains a value representing an integer.
Example: root = [5, 3, 8, 1, 4, null, 9]
Constraints:
• 2 <= number of nodes <= 5000
• 0 <= Node.val <= 105
Output: The output should be a single integer representing the maximum absolute difference between the values of two nodes, where one is an ancestor of the other.
Example: Output: 8
Constraints:
• The absolute difference is calculated between two nodes, where one node is an ancestor of the other.
Goal: The goal is to find the maximum difference between values of two nodes in a binary tree where one is an ancestor of the other. This can be done by recursively traversing the tree and tracking the minimum and maximum values encountered along the path.
Steps:
• 1. Start by recursively visiting each node of the tree.
• 2. Track the minimum and maximum values encountered on the path from the root to the current node.
• 3. Calculate the difference between the current node's value and the tracked minimum and maximum values.
• 4. Return the maximum difference found.
Goal: Ensure the solution is efficient given the constraints of the problem.
Steps:
• The solution must handle a tree with up to 5000 nodes.
• Node values must be between 0 and 105.
Assumptions:
• The tree is a valid binary tree with at least two nodes.
• Each node has a value between 0 and 105.
• Input: Input: root = [5, 3, 8, 1, 4, null, 9]
• Explanation: In this example, the maximum difference occurs between nodes 5 and 1, yielding a difference of 8. Hence, the output is 8.
• Input: Input: root = [10, 5, 15, 3, 7, null, 20]
• Explanation: In this case, the maximum difference is between nodes 10 and 3, yielding a difference of 7. Thus, the output is 7.
Approach: The approach uses a depth-first search (DFS) strategy to traverse the tree while keeping track of the maximum and minimum values encountered along the path from the root to each node.
Observations:
• The problem involves traversing a binary tree and calculating the maximum and minimum differences for every path.
• A depth-first search approach is appropriate because it allows us to visit all nodes while keeping track of necessary values (min and max).
Steps:
• 1. Perform a DFS traversal on the tree, starting from the root.
• 2. At each node, compute the difference between its value and the tracked minimum and maximum values.
• 3. Propagate the minimum and maximum values down the tree as you visit the child nodes.
• 4. At each node, compute the maximum of the difference between the node's value and both the minimum and maximum values encountered so far.
• 5. Return the maximum difference after visiting all nodes.
Empty Inputs:
• If the tree has fewer than two nodes, return 0 (this case is assumed to be invalid per problem constraints).
Large Inputs:
• The algorithm should efficiently handle trees with up to 5000 nodes.
Special Values:
• Ensure that the solution handles cases where all node values are the same (resulting in a difference of 0).
Constraints:
• The tree must be valid, and each node must have a value between 0 and 105.
int maxAncestorDiff(TreeNode* root) {
return helper(root, root->val, root->val);
}
int helper(TreeNode* node, int mx, int mn) {
if(!node) return mx - mn;
mx = max(mx, node->val);
mn = min(mn, node->val);
return max(helper(node->left, mx, mn), helper(node->right, mx, mn));
}
1 : Function Definition
int maxAncestorDiff(TreeNode* root) {
Define the function `maxAncestorDiff` which takes the root of the tree and calculates the maximum ancestor-descendant difference.
2 : Function Call
return helper(root, root->val, root->val);
Call the helper function with the root node and initialize both the maximum (`mx`) and minimum (`mn`) values to the root's value.
3 : Helper Function Definition
int helper(TreeNode* node, int mx, int mn) {
Define the helper function which recursively traverses the tree, updating the maximum and minimum values and computing the maximum difference.
4 : Base Case
if(!node) return mx - mn;
If the current node is null, return the difference between the maximum and minimum values encountered so far.
5 : Update Maximum
mx = max(mx, node->val);
Update the maximum value (`mx`) with the current node's value if it is larger.
6 : Update Minimum
mn = min(mn, node->val);
Update the minimum value (`mn`) with the current node's value if it is smaller.
7 : Recursive Traversal
return max(helper(node->left, mx, mn), helper(node->right, mx, mn));
Recursively calculate the maximum difference for the left and right subtrees, and return the maximum of the two results.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n), where n is the number of nodes in the tree, since we visit each node once during the DFS traversal.
Best Case: O(1)
Worst Case: O(h)
Description: The space complexity is O(h), where h is the height of the tree, due to the space used by the recursion stack. In the worst case (unbalanced tree), h can be O(n).
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